There are two types of reciprocating compressors in the market: single stage (shown in FIG. 1) and multi stage compressors (for example the double stage compressor shown in FIG. 2). The working pressure of single stage compressors is limited to under 150 psi. If higher working pressure is required, as in many heavy duty applications, then switching to double or multi stage compressors is inevitable.
FIG. 2 illustrates a typical double stage reciprocating compressor. In double stage compression strategy, the compression process is broken up into two stages. In the first stage, the cylinder receives the fresh air at atmospheric pressure and compresses it using a piston. The compressed air is urged into a low pressure air tank (an intercooler) where some of the heat produced by compression is removed. The air is then channelled to a second cylinder, where it is compressed further to the desired pressure. The air is then channelled to a high pressure air tank for storage. Since the double stage compressors consist of a minimum of two cylinders, they weigh more than single stage compressors. They also have higher energy loss due to the higher piston cylinder friction.
Meanwhile, the automotive industry has seen itself in a marathon of advancement during the last decade. This is partly due to the global environmental concerns on the increase of air pollution and decrease of fossil fuel resources. The next generation of vehicles must be cleaner and more efficient than the current conventional ones. To this end, vehicle manufacturers have tried different innovations: pure electric, fuel cell and hybrid electric vehicles. The pure electric and fuel cell vehicles have not yet proven to be a convenient solution to environmental problems. Compared to conventional vehicles, the traveling range of pure electric vehicles is very low due to the use of batteries, which provide a limited source of energy. On the other hand, it has not yet been possible to commercialize fuel cell technology.
Hybrid electric vehicles have overcome the production limits of pure electric and fuel cell vehicles and are regarded as one of the most effective and feasible solution to environmental concerns. Despite the beneficial improvements that this kind of vehicle provides, there are some serious concerns about their high manufacturing price, complexity and limited battery life.
Typical air hybrid engines operate similarly to typical hybrid electric engines. FIG. 3 illustrates the interconnection between components of a typical air hybrid engine. The air hybrid engine uses two energy sources, fuel and pressurized air. The air hybrid engine absorbs a vehicle's kinetic energy while braking and stores it in the form of compressed air to a storage tank. The compressed air is then used while accelerating. The air is compressed using a single stage compression approach.
FIG. 4 illustrates a cross-sectional view of one example of an air hybrid engine. A typical air hybrid engine has an extra valve per cylinder relative to a traditional four stroke engine, which connects each cylinder to the air tank. During braking, this extra valve opens and the exhaust valve closes, allowing the engine to work as an air compressor, charging the air tank with high pressure air. This pressurized air can later be used to drive the internal combustion engine as an air motor, or it can be used in the combustion process during high energy demand leading to a higher efficiency relative to a fuel-only drive.
Air hybrid engines are typically more efficient than conventional engines because they recover the vehicle's kinetic energy while braking, reduce fuel consumption during a cold start, and enable the engine to work with higher pressure than conventional engines.
A typical air hybrid engine has five modes, namely the compression mode, the air motor mode, air power assisted mode (supercharged) and combustion (conventional) and start up mode.
The compression mode is illustrated in FIG. 5. This mode is activated when the driver applies the brake pedal. In this mode, fuel is shut off and the engine works as a two stroke air compressor and the piston compresses the air into the air tank while the exhaust valve remains deactivated, storing the vehicle's kinetic energy in the shape of pressurized air in the air tank.
The air motor mode is shown in the FIG. 6. The valve between the air tank and the cylinder opens, allowing the pressurized air to run the engine as a two stroke air motor. This mode is activated when the power demand is low or at cold start to avoid high fuel consumption.
The air power assisted mode (supercharged) is shown in FIG. 7 and is activated when the desired torque is high. The intake valve is deactivated and pressurized air is delivered from the air tank leading to a more efficient combustion process in the cylinder. The engine is provided with pressurized air from the air tank instead of from the atmosphere. The mass of fuel and air entering the engine cylinders is increased, which in turn increases the produced power significantly in this mode. In contrast to typical supercharged engines which have lower efficiency at low speeds and loads, air hybrid engines can be supercharged at any operating point thanks to stored air in the air tank. Conventional mode is also activated when the desired load is moderate or the air tank pressure is relatively low or empty. The stored air in the air tank can also be used to run the engine at cold start. This mode is the start up mode.
In the combustion mode, the air tank valve is closed while the intake and exhaust valves are used for enabling driving of the engine as a typical four stroke engine.
As is commonly known, in typical city driving (where stop and go driving is common) a significant fraction of energy is consumed in braking. For instance, in EPA FTP75 urban driving cycle approximately 40% of the energy is wasted while braking. Thus, if the braking system can recover the braking energy, the vehicle energy consumption will be reduced significantly. Air hybrid engines have been developed to capture and store the braking energy for further use.
The ideal air cycle of the single tank system is shown in FIG. 8. When the piston is at the Bottom Dead Center (BDC), the intake valve closes. The piston starts moving up to the Top Dead Center (TDC) and compresses the air adiabatically.
The charging valve opens when the air pressure in the cylinder equals the tank pressure. At this time, air enters the tank in a constant pressure process, assuming that the air tank is big enough and its pressure does not change while charging. The charging valve closes when the piston is at TDC. The piston moves down and the intake valve opens when the pressure in the cylinder equals the atmospheric pressure. The aforementioned cycle is the ideal cycle and has the highest stored air mass in the air tank to the consumed energy ratio comparing to any other cycle.
The maximum amount of air mass that can be stored in the air tank is limited, based on the following relation:
                                          m            max                    =                                    C              r                        ⁢                                          P                atm                                                              T                  atm                                ⁢                R                                      ⁢                          V              tank                        ⁢            M                          ,                            (        1        )            where R is the ideal gas constant, Vtank is the air tank volume, M is the air molecular mass, and Cr is the cylinder compression ratio. Setting the maximum allowable temperature of the air tank, its maximum pressure also can be defined based on the above equation. By increasing the cylinder compression ratio, the capacity of energy storing can be increased, however this will result in higher temperature which deteriorates the efficiency of the system.
The above relation can be proven with reference to FIG. 8. Suppose that the air tank is already full and its pressure and temperature are Ptank and Ttank. Air tank pressure and temperature are related based on equation (1), by the following relation:
                              P          tank                =                                                            P                atm                            ⁢                              T                tank                                                    T              atm                                ⁢                      C            r                                              (        2        )            
At point 1, the air mass inside the cylinder is:
                              m          1                =                                                            P                atm                            ⁢                              V                cyl                                                    RT              atm                                ⁢          M                                    (        3        )            
Considering adiabatic compression and ideal mixing of gases, cylinder pressure at the arbitrary point 2 is:
                              P          2                =                                                                                                  P                    atm                                    ⁡                                      (                                                                  V                        cyl                                                                    V                        *                                                              )                                                  k                            ⁢                              V                *                                      +                                                                                P                    atm                                    ⁢                                      T                    tank                                                                    T                  atm                                            ⁢                              V                tank                            ⁢                              C                r                                                                        V              *                        +                          V              tank                                                          (        4        )            and the temperature at point 2 is
                              T          2                =                                                                                                  P                    atm                                    ⁡                                      (                                                                  V                        cyl                                                                    V                        *                                                              )                                                  k                            ⁢                              V                *                                      +                                                                                P                    atm                                    ⁢                                      T                    tank                                                                    T                  atm                                            ⁢                              V                tank                            ⁢                              C                r                                                                                                                                                    P                      atm                                        ⁡                                          (                                                                        V                          cyl                                                                          V                          *                                                                    )                                                        k                                ⁢                                  V                  *                                                                                                  T                    atm                                    ⁡                                      (                                                                  V                        cyl                                                                    V                        *                                                              )                                                                    k                  -                  1                                                      +                                                            P                  atm                                                  T                  atm                                            ⁢                              V                tank                            ⁢                              C                r                                                                        (        5        )            
Air pressure and temperature at point 3 are defined by equations (6) and (7).
                              P          3                =                                            P              2                        (                                                            V                  tank                                +                                  V                  *                                                                              V                  tank                                +                                                      V                    cyl                                                        C                    r                                                                        )                    k                                    (        6        )                                          T          3                =                                            T              2                        (                                                            V                  tank                                +                                  V                  *                                                                              V                  tank                                +                                                      V                    cyl                                                        C                    r                                                                        )                                k            -            1                                              (        7        )            
The charging valve closes at point 3 so the amount of air mass trapped in the cylinder dead volume can be found as follows:
                              m          trapped                =                                                            P                3                            ⁢                                                V                  cyl                                                  C                  r                                                                    RT              3                                ⁢          M                                    (        8        )            
By plugging equations (6) and (7) into equation (8), the trapped mass in the cylinder dead volume becomes:
                              m          trapped                =                                                            P                atm                            ⁢                              V                cyl                                                    RT              atm                                ⁢          M                                    (        9        )            equaling the amount of air mass entered into the cylinder at point ‘1’. This proves that the maximum amount of air mass in the air tank is limited by equation (1).
The above mentioned braking cycle can be used to model regenerative braking, as illustrated in FIG. 9 of a typical air hybrid engine vehicle with the specification shown in Table 1, which models a 1400 kg vehicle decelerating from 90 km/hr to 10 km/hr using only regenerative braking.
TABLE 1Vehicle Mass1400 kgVehicle Initial Velocity90 km/hrVehicle Final Velocity10 km/hrTransmition Ratio5.7Cylinder Volume2 LAir Tank Volume30 LAir Tank Temperature750KAir Tank Initial Pressure1 barCompression Ratio10
FIG. 10 illustrates the pressure profile in the air tank versus time for a typical air hybrid engine implementation. As can be seen, the pressure in the storage increases but there is a limit for the pressure in the air tank. In a particular implementation, the pressure in the air tank builds up to 25 bar but it cannot go further beyond this value. Furthermore, the efficiency of regenerative braking is limited in this implementation to about 22% and the braking time (using only regenerative braking) is about 17.1 s.
Capturing 22% of the vehicle's kinetic energy is significant, however storage could be improved to enhance efficiency. There are two options to increase the capacity of energy storing in the air tank, either using a higher volume tank or increasing the pressure. Increasing the volume of the tank is not a viable solution due to the lack of the space in the vehicle. On the other hand, increasing the pressure is not achievable in current air hybrids because, the maximum pressure is limited by the engine compression ratio.
Furthermore, in contrast with conventional engines which have only one mode of operation (combustion), air hybrid engines have five modes of operation as described above. At each mode, a different type of cycle should be followed, with each cycle having different valve timing. Thus a camless valvetrain is typically required for air hybrid engine control.
A conventional valvetrain limits the performance of an engine but has more operational advantages over a camless valvetrain because valve motion is governed by the cam profile, which is typically designed to have low seating velocity. Seating velocity in the camshaft design is limited below 0.5 m/s. The valve's low seating velocity leads to durability and low noise. In contrast, a typical camless valvetrain, which has no mechanical connection with engine, introduces a difficult control problem. Control techniques should be applied to perform both accurate valve timing and low seating velocity [4, 7]. This introduces a very complicated problem, especially in the case of an air hybrid engine, in which the valve timing changes to compensate for different desired loads. The controller therefore must be robust enough to account for engine speed, tank pressure and desired torque variations.
What is required, therefore, is a method for more optimally compressing air. What is also required is an air hybrid engine operable to more optimally compress air than current air hybrid engines. A more optimal camless valvetrain would also be beneficial for controlling air hybrid engines.